4u Mathematics Marathon V 1.0 (1 Viewer)

[EmmR]

Back in 2005 there was some form of question numbering system but it seemed to dissapear at Q4.

what question number are you guys up to? Q76?


So whats the next question?

 

[pLuvia]

Q: If I_n = ∫₀¹(1-x²)ⁿ dx, find I_n in terms of I_n-1 and n.

Spoiler

int.{0 to 1}(1-x²)ⁿ dx
Let u=(1-x²)ⁿdx dv=dx
du=-2xn(1-x²)^n-1dx v=x
I_n
=int.{0 to 1}(1-x²)ⁿ dx
=[x(1-x²)ⁿ]{0 to 1}+int.{0 to 1}n2x²(1-x²)^n-1dx
=2n int.{0 to 1}x²(1-x²)^n-1dx
=2n int.{0 to 1}-(1-x²-1)(1-x²)^n-1dx
=-2n int.{0 to 1}(1-x²)ⁿ-(1-x²)^n-1dx
=-2n[I_n-I_n-1]
I_n(1+2n)=2nI_n-1
I_n=2nI_n-1/(2n+1)

 

[pLuvia]

Opps, I'll edit it
Yeh forgot to put it in after the first IBP

 

[pLuvia]

Just put an easy one to get this thread bumped up again

Prove that
[a+b+c]/3>√abc

 

[Riviet]

Any restrictions on a, b and c?

 

[pLuvia]

a,b,c are real positive numbers

 

[pLuvia]

Prove a/b + b/c + c/a + c/b + b/a + a/c > 6

Spoiler

a/b+b/c+c/a+c/b+b/a+a/c-6
=[a²c+ab²+bc²+ac²+b²c+a²b-6abc]/abc
=[1/abc][a(b²-2bc+c²)+b(a²-2ac+c²)+c(a²-2ab+b²)]
=[1/abc][a(b-c)²+b(a-c)²+c(a-b)²]
>0
Hence a/b+b/c+c/a+c/b+b/a+a/c>6

 

[Mountain.Dew]

Spoiler

We know that a² + b² > 2ab.

Therefore a + b > 2√ab by substitution.

and similarly b + c > 2√bc and a + c > 2√ac

So if we add all these together we get 2 (a + b + c) > 6√abc

Therefore a + b + c > 3√abc.

whats the 'proof' for the fact that 2√ac + 2√bc + 2√ac > 6√abc?

is it assumed that √ac or √bc or √ab > 1?

Prove a/b + b/c + c/a + c/b + b/a + a/c > 6

Spoiler

a/b+b/c+c/a+c/b+b/a+a/c-6
=[a²c+ab²+bc²+ac²+b²c+a²b-6abc]/abc
=[1/abc][a(b²-2bc+c²)+b(a²-2ac+c²)+c(a²-2ab+b²)]
=[1/abc][a(b-c)²+b(a-c)²+c(a-b)²]
>0
Hence a/b+b/c+c/a+c/b+b/a+a/c>6

ooooo very elegant solution

 

[pLuvia]

ooooo very elegant solution

What's the other way of doing it? Just wanting to know

 

[pLuvia]

Extension to the question
First prove that [a+b+c+d]/4=⁴√abcd
Then find an expression for d in terms of a,b and c

That should help

 

[Yip]

(a+b)/2>√ab
(c+d)/2>√cd
(a+b+c+d)/2>√ab+√cd
a+b+c+d>2(√ab+√cd)>2(2√√abcd)=4⁴√abcd
[a+b+c+d]/4=⁴√abcd

Letting d=a+b+c/3,

LHS=4/3(a+b+c)>4⁴√abc(⁴√[a+b+c]/⁴√3)
[(a+b+c)^(3/4)]/3>⁴√abc/⁴√3
[(a+b+c)^(3/4)]>⁴√27abc
(a+b+c)^3>27abc
a+b+c>3³√abc

Prove that
[a+b+c]/3>√abc

I think pluvia had a typo,
if u sub a=1,b=2,c=3, u get 2>√6 which is obviously false
2>³√6 makes more sense

UMAT tomorrow -_- nande...~.~

 

[Mountain.Dew]

What's the other way of doing it? Just wanting to know

we know that:

a² + b² > 2ab

therefore, when divide a/b + b/a > 2

hence, b/c + c/b > 2, c/a + a/c > 2

add em all together: a/b + b/c + c/a + c/b + b/a + a/c > 6

ummm i take back my comment on your 'elegant solution'. i find this far easier and more efficient sorry, tis was my bad, pLuvia.

 

[Riviet]

Ooh nice proof Yip, also good luck with UMAT too, don't stay up too late studying for it!

 

[pLuvia]

Opps my bad, I actually meant to type
[a+b+c/3]³>√abc
Sorry LottoX

Good luck with the UMAT

 

[bboyelement]

S'ok, as long as I know it's not my crapiness making me fail the question.

I can't believe the UMAT is actually tomorrow. I feel so nervous.

Anyway:

Given: t/(a+b) = 1, prove 1/a² + 1/b² > 8/t²

therefore t=a+b
1/a²+ 1/b² = (1/a - 1/b)² + 2/ab
1/a²+ 1/b² > 2/ab
(a + b)² = (a - b)²+ 4ab
(a + b)² > 4ab
ab< (a + b)²/4

1/a² + 1/b² > 2/(a + b)²/4
1/a² + 1/b² > 8/(a + b)²
1/a² + 1/b² > 8/t²

prove
(b + c + d)(a + c + d)(a + b + d)(a + b + c) > 81abcd

btw. a,b,c,d all bigger than 0

 

[Riviet]

prove
(b + c + d)(a + c + d)(a + b + d)(a + b + c) > 81abcd

btw. a,b,c,d all bigger than 0

By AM/GM inequality:
(b+c+d)/3 > (bcd)^1/3 (1)

Similarly,
(a+c+d)/3 > (acd)^1/3 (2)

(a+b+d)/3 > (abd)^1/3 (3)

(a+b+c)/3 > (abc)^1/3 (4)

(1)(2)(3)(4) => {(b+c+d)(a+c+d)(a+b+d)(a+b+c)}/81 > (a³b³c³d³)^1/3

.'. (b+c+d)(a+c+d)(a+b+d)(a+b+c) > 81abcd, as required.

 

[Riviet]

Next Question:
The region bounded by y=1/(x+1), y=x/(x+1) and the y-axis is rotated about the y-axis through one complete revolution.

i) Use the method of cylindrical shells to show that the volume of the solid of revolution is given by:
aaaaaa1
V = 2pi∫{x(1-x)/(1+x)} dx
aaaaa0

ii) Write x(1-x)/(1+x) in the form ax + b + c/(x+1) to find the exact value of the above integral and hence the volume of the solid of revolution.

 

[Riviet]

= 2π{-x²/2 + 2x - 2ln(x+1)}₀¹

= 2π(1/2 + 2 - 2ln2)

Almost had it.

Don't forget a new question.

 

[Mountain.Dew]

Bah, I forgot both. Anyway:

Show that for n > 1

1ln(2/1) + 2ln(3/2) + ... + nln{(n+1)/n} = ln({n+1}ⁿ/n!)

1ln(2/1) + 2ln(3/2) + ... + nln{(n+1)/n}

LHS = ln 2 - ln 1 + 2ln3 - 2ln2 + 3ln4 - 3ln3 + 4ln5 - 4ln4 + ... + (n-1)ln(n) - (n-1)ln(n-1) + nln(n+1) - nln(n) = - ln1 - ln2 - ln3 - ln4 - .... - ln(n) + nln(n+1)

= nln(n+1) - (ln1 + ln2 + ln3 +...+ln(n)) = nln(n+1) - ln(1.2.3.4...n) = ln(n+1)ⁿ - ln(n!) = ln({n+1}ⁿ/n!) = RHS

 

[Mountain.Dew]

New question: tis an easy one

If a, b, c are roots of x^3 -9x + 9 = 0, show that

(a-1)(b-1)(c-1) = -1

4 marks.